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Question Number 14564 by tawa tawa last updated on 02/Jun/17

What is the last 2 digits of 2^(613)

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{last}\:\mathrm{2}\:\mathrm{digits}\:\mathrm{of}\:\:\:\:\:\:\mathrm{2}^{\mathrm{613}} \\ $$

Commented by tawa tawa last updated on 02/Jun/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by tawa tawa last updated on 02/Jun/17

But the question is last two digit

$$\mathrm{But}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{last}\:\mathrm{two}\:\mathrm{digit} \\ $$

Commented by Tinkutara last updated on 02/Jun/17

2 ≡ 2 (mod 100) 2^2 ≡ 4 (mod 100) 2^4 ≡ 16 (mod 100) 2^8 ≡ −44 (mod 100) 2^(16) ≡ 36 (mod 100) 2^(32) ≡ −4 (mod 100) 2^(64) ≡ 16 (mod 100) 2^(128) ≡ −44 (mod 100) 2^(256) ≡ 36 (mod 100) 2^(512) ≡ −4 (mod 100) 2^(613) = 2^(512 + 64 + 32 + 4 + 1) = (−4)(16)(−4)(16)(2) = 8192 ≡ 92 (mod 100) Hence last two digits are 92.

$$\mathrm{2}\:\equiv\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{100}\right) \\ $$$$\mathrm{2}^{\mathrm{2}} \:\equiv\:\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{100}\right) \\ $$$$\mathrm{2}^{\mathrm{4}} \:\equiv\:\mathrm{16}\:\left(\mathrm{mod}\:\mathrm{100}\right) \\ $$$$\mathrm{2}^{\mathrm{8}} \:\equiv\:−\mathrm{44}\:\left(\mathrm{mod}\:\mathrm{100}\right) \\ $$$$\mathrm{2}^{\mathrm{16}} \:\equiv\:\mathrm{36}\:\left(\mathrm{mod}\:\mathrm{100}\right) \\ $$$$\mathrm{2}^{\mathrm{32}} \:\equiv\:−\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{100}\right) \\ $$$$\mathrm{2}^{\mathrm{64}} \:\equiv\:\mathrm{16}\:\left(\mathrm{mod}\:\mathrm{100}\right) \\ $$$$\mathrm{2}^{\mathrm{128}} \:\equiv\:−\mathrm{44}\:\left(\mathrm{mod}\:\mathrm{100}\right) \\ $$$$\mathrm{2}^{\mathrm{256}} \:\equiv\:\mathrm{36}\:\left(\mathrm{mod}\:\mathrm{100}\right) \\ $$$$\mathrm{2}^{\mathrm{512}} \:\equiv\:−\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{100}\right) \\ $$$$\mathrm{2}^{\mathrm{613}} \:=\:\mathrm{2}^{\mathrm{512}\:+\:\mathrm{64}\:+\:\mathrm{32}\:+\:\mathrm{4}\:+\:\mathrm{1}} \\ $$$$=\:\left(−\mathrm{4}\right)\left(\mathrm{16}\right)\left(−\mathrm{4}\right)\left(\mathrm{16}\right)\left(\mathrm{2}\right)\:=\:\mathrm{8192} \\ $$$$\equiv\:\mathrm{92}\:\left(\mathrm{mod}\:\mathrm{100}\right) \\ $$$$\mathrm{Hence}\:\mathrm{last}\:\mathrm{two}\:\mathrm{digits}\:\mathrm{are}\:\mathrm{92}. \\ $$

Commented by Tinkutara last updated on 02/Jun/17

2 ≡ 2 (mod 1000) 2^2 ≡ 4 (mod 1000) 2^4 ≡ 16 (mod 1000) 2^8 ≡ 256 (mod 1000) 2^(16) ≡ −464 (mod 1000) 2^(32) ≡ 296 (mod 1000) 2^(64) ≡ −384 (mod 1000) 2^(128) ≡ 456 (mod 1000) 2^(256) ≡ −64 (mod 1000) 2^(512) ≡ 96 (mod 1000) 2^(613) = 2^(512 + 64 + 32 + 4 + 1) = (96)(−384)(296)(16)(2) ≡ −808 (mod 1000) ≡ 192 (mod 1000) Last 3 digits are 192.

$$\mathrm{2}\:\equiv\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{1000}\right) \\ $$$$\mathrm{2}^{\mathrm{2}} \:\equiv\:\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{1000}\right) \\ $$$$\mathrm{2}^{\mathrm{4}} \:\equiv\:\mathrm{16}\:\left(\mathrm{mod}\:\mathrm{1000}\right) \\ $$$$\mathrm{2}^{\mathrm{8}} \:\equiv\:\mathrm{256}\:\left(\mathrm{mod}\:\mathrm{1000}\right) \\ $$$$\mathrm{2}^{\mathrm{16}} \:\equiv\:−\mathrm{464}\:\left(\mathrm{mod}\:\mathrm{1000}\right) \\ $$$$\mathrm{2}^{\mathrm{32}} \:\equiv\:\mathrm{296}\:\left(\mathrm{mod}\:\mathrm{1000}\right) \\ $$$$\mathrm{2}^{\mathrm{64}} \:\equiv\:−\mathrm{384}\:\left(\mathrm{mod}\:\mathrm{1000}\right) \\ $$$$\mathrm{2}^{\mathrm{128}} \:\equiv\:\mathrm{456}\:\left(\mathrm{mod}\:\mathrm{1000}\right) \\ $$$$\mathrm{2}^{\mathrm{256}} \:\equiv\:−\mathrm{64}\:\left(\mathrm{mod}\:\mathrm{1000}\right) \\ $$$$\mathrm{2}^{\mathrm{512}} \:\equiv\:\mathrm{96}\:\left(\mathrm{mod}\:\mathrm{1000}\right) \\ $$$$\mathrm{2}^{\mathrm{613}} \:=\:\mathrm{2}^{\mathrm{512}\:+\:\mathrm{64}\:+\:\mathrm{32}\:+\:\mathrm{4}\:+\:\mathrm{1}} \\ $$$$=\:\left(\mathrm{96}\right)\left(−\mathrm{384}\right)\left(\mathrm{296}\right)\left(\mathrm{16}\right)\left(\mathrm{2}\right) \\ $$$$\equiv\:−\mathrm{808}\:\left(\mathrm{mod}\:\mathrm{1000}\right) \\ $$$$\equiv\:\mathrm{192}\:\left(\mathrm{mod}\:\mathrm{1000}\right) \\ $$$$\mathrm{Last}\:\mathrm{3}\:\mathrm{digits}\:\mathrm{are}\:\mathrm{192}. \\ $$

Commented by tawa tawa last updated on 02/Jun/17

i really appreciate sir.

$$\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$

Commented by mrW1 last updated on 02/Jun/17

good working! pls find the last 3 digits.

$${good}\:{working}! \\ $$$${pls}\:{find}\:{the}\:{last}\:\mathrm{3}\:{digits}. \\ $$

Commented by Tinkutara last updated on 02/Jun/17

I do not know how to find last 3 digits. Maybe in that case, mod 1000 is required. It will be very lengthy and calculative (use of calculators).

$$\mathrm{I}\:\mathrm{do}\:\mathrm{not}\:\mathrm{know}\:\mathrm{how}\:\mathrm{to}\:\mathrm{find}\:\mathrm{last}\:\mathrm{3}\:\mathrm{digits}. \\ $$$$\mathrm{Maybe}\:\mathrm{in}\:\mathrm{that}\:\mathrm{case},\:\mathrm{mod}\:\mathrm{1000}\:\mathrm{is} \\ $$$$\mathrm{required}.\:\mathrm{It}\:\mathrm{will}\:\mathrm{be}\:\mathrm{very}\:\mathrm{lengthy}\:\mathrm{and} \\ $$$$\mathrm{calculative}\:\left(\mathrm{use}\:\mathrm{of}\:\mathrm{calculators}\right). \\ $$

Commented by RasheedSoomro last updated on 02/Jun/17

e^x cellent!

$$\mathrm{e}^{\mathrm{x}} \mathrm{cellent}! \\ $$

Commented by mrW1 last updated on 02/Jun/17

very good! please try to solve Q13724

$${very}\:{good}! \\ $$$${please}\:{try}\:{to}\:{solve} \\ $$$${Q}\mathrm{13724} \\ $$

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